Some aspects of the Jacobian conjecture: (the geometry of automorphisms of ℂ²)

(THE GEOMETRY OF AUTOMORPHISMS OF

ℂ

²)

A. Hamid A. Hussain Ali

A Thesis Submitted for the Degree of PhD

at the

University of St Andrews

1987

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(THE GEOMETRY OF AUTOMORPHISMS OF

)

BY

A

A

We consider the affine varieties which arise by 9

considering invertible polynomial maps from C to itself of less than or equal to a given-dtegree. These varieties arise naturally in the investigation of the long-standing Jacobian Conjecture.

We start with some calculations in the lower degree cases. These calculations provide a proof of the Jacobian conjecture in these cases and suggest how the investigation in the higher degree cases should proceed.

We then show how invertible polynomial maps can be decomposed as products of what we call triangular maps and we are able to prove a uniqueness result which gives a stronger version of Jung's theorem [j] which is one of the most important results in this area. Our proof also gives a new derivation of Jung's theorem from Segre's lemma.

We give a different decomposition of an invertible polynomial map as a composition of "irreducible maps" and we are able to write down standard forms for these irreducibles. We use these standard forms to give a description of the structure of the varieties of invertible m a p s .

To my wife Khawla, my | sons Khazraj, Hassan and

Barra and my relatives in Iraq.

i

%

CERTIFICATE

I A Hamid A Hussain Ali hereby certify that this thesis has been composed by myself, that it is a record of my own work, and that it has not been accepted in partial or complete fulfilment of any other degree or professional qualification.

I was admitted to the Faculty of Science of the University of St Andrews under Ordinance General No 12 in October 1982 and as a candidate for the degree of PhD in February 1984.

Signed Date . . A /. ( ^ Â 7. . . ,

I hereby certify that the candidate has fulfilled the conditions of the Resolution and Regulations appropriate to the Degree of PhD.

ACKNOWLEDGEMENTS

I take this opportunity to express my gratitude to Dr J J O'Connor, my supervisor for suggesting the project and for his kind advice, stimulating discussions and encouragement throughout my study period.

It is my pleasure to record sincere appreciation and gratitude to my wife, Khawla for her patience and

encouragment throughout this study.

COPYRIGHT DECLARATION

In submitting this thesis to the University of St Andrews I understand that I am giving permission for it to be made available for public use in accordance with the regulations of the University Library for the time being in force, subject to any copyright vested in the work not being affected thereby. I also understand that the title and abstract will be published, and that a copy of the work may be made and supplied to any bona fide library or research worker.

Dedication

V

CONTENTS

Pages

Certificate ₁₁

Acknowledgements 111

Copyright Declaration _IV

Contents V

Chapter One: Introduction

Chapter Two:

1 - 2

Calculations for invertible 3 - 2 9 polynomial maps of low degree.

;s y

Chapter Three: General results on maps with 30 - 42 Constant Jacobian.

Chapter Four:

Chapter Five:

Standard forms for invertible 43 - 55 polynomial maps from to

itself.

Spaces and Varieties for invert- 5 6 - 7 1 ible polynomial maps»

Chapter Six: Elements of finite order in the space of invertible polynomial m a p s .

72 - 83

Conclusion 84 - 88

INTRODUCTION

A mapping f:<I^---> f(x) = (f^(x), ... ,f^(x)), is a polynomial mapping if each f^ is a polynomial.

- 1

If f is invertible (ie if f is a polynomial map), then since the Jacobian satisfies J(hog) = Jh(g).Jg, Jf is invertible also and hence (since the determinant of J.f is a polynomial) det (Jf) must be a non-zero constant poly­ nomial. The converse of this result was first conjectured by Keller [K] in 1 939 and this is known as the Jacobian Conjecture.

Much evidence has been assembled in favour of this conjecture (see the article by Bass, Connell and Wright [BCW]), but even for the case n = 2, the conjecture is. still open. (Though in this case Moh shows that it holds for all

2 2

maps from Πto (T with degree less than or equal to 100 [Mo]).

If f is a general polynomial map of given degree, then the equation;

det (Jf) = Constant 0 ]

1

can be regarded as a set of simultaneous equations in the

coefficients of f and hence defines an affine variety in the j

n -i

In this work, we study the case n = 2.

CALCULATIONS FOR INVERTIBLE POLYNOMIAL MAPS OF LOW DEGREE I’

--- r

In this chapter, the polynomial maps from to itself # o

of degrees 1, 2, 3 and 4 which map (0,0)e CE to itself are considered. (Later one can see proofs for some general results which avoid the necessity for some of these intricate calculations.)

Any polynomial map can be formed from the composition of an origin-preserving map with a translation map and from

now on we assume that all maps are origin preserving.

The set of all origin-preserving linear maps â (x,y) 1----^ (a^ X + ^ y, ^ x + ^ y)

forms a space of dimension 4 where this map is identified with the point:

(a^, a^ , , b^ ) e CE'*

(This identification is the standard Cremona map. )

For such maps the Jacobian Conjecture is true trivially and the set of all such invertible maps is an open subset of CE"^ given by:

det Jf ^ 0 or a^ b2 - &2 ^ 0

and hence is an affine variety of dimension 4. So the first non-trivial case is the set of polynomial maps of degree - 2

2

Theorem Any Polynomial map from CE to itself of degree - 2 whose Jacobian determinant is a non-zero constant is invertible and is of the form:

s b. - ka^

t “ b_ - ka. ®4 *^5 ^ ®5 *^4

The variety of all such maps is 6 dimensional

Proof A typical origin-preserving polynomial map of degree - 2 is of the form:

2 2

f(x,y) = (a.j X + a^ xy + a^ y + a. x + a^ y,

2 2

b.j X + bg xy t bg y + b^ x + b^ y

and so the set of all such maps forms a space of dimension

1 0

.

The condition det Jf = constant ^ 0 is thus: 2a. X + a^ y + a^ a^ x + 2a^ y + a^

|Jf|

bg X + 2b^ y + b^ 2b.| X + bg y t b^

By equating coefficients of powers x and y, this is equivalent to the following set of 5 equations and one

inequality:

X : a^ bg = b^ a^

x y : a^ b = b^ a^

y: 2b3 b^ = 2b^ + b^

Ijfl = a. b_ - 3g b. = constant ^ 0 |

i

One can solve these equations to find a general form for a map of degree ^ 2 with a constant determinant as follows :

One can assume a^ ^ 0 (One of the 2nd order terms must be present. If, for example we have a.^ = 0 and a^ ^ 0, then compose on the right with the map (x,y)|— >(x, e x + y) and we get a map with the required term for a suitable choice of e) .

Let b- = ka. , then k = — .- We now equate coefficients of_{I I a .j} terms in the Jacobian determinant to zero.

2 ^1

Coeeficients of x : bg = — _{6 a z} ~ ka.. _z, %

of xy: bg = — 3 3 = ka3

proportional.

Let g(x,y) = (x, kx - y) and then

2

gof = (a,| X t .... + a^ x + a^ y, (ka^ - ) x + (ka^ - b^ ) y) We can use a "Tschirnhaus" transformation to get rid of the term in xy. Namely let

^2

h (x,y) = (x - 2^ y, y) and then

2

9 4a. a_ - a„ „ 2a. a^. - a« a. gofoh = (a ,

y

b,j a ^ ~ a ^ b ^ 2a,| (b.j a ^ a ^ b ^ ) a2(b^ a ^ a .j b^ )

From the Jacobian, equating coefficients;

2

y: (b,j a^ - a^ b^ ) (4a^ - &2 ) = 0

x; 2a^ (b,j a^ - a^^ b^ ) = a^fb^ a^ - a^ b^ )

If b\| a^ - a^ b^ = 0, then we would have b,^ a^ = a^ bg (since a^ ^ 0) and then b^ = ka^ and b^ = ka^ , which would lead to

I

I

^2 b, a^ - a, b^

Le &2 is determined by: (a^, b ^ , a ^ , b ^ , a^, b ^ )

2

Also 4a,| a^ - ^2 “ ® Hence

gofoh = { a ^ + a X + — ---t l - . . ^ A x),

2ai a,

then f = g~''o{a, x^+ a^x + , i l L i --- ! l..i ix)oh~''

b. g 2a. ar — a^ a. f = (X, — X - y)o(a, X + a^ X +— ---y

i l L l L L l i i x)o( X t ! t . y , y ) 2a^

So f can be written as a composition:

(Linear map) o (Triangular map) o (Linear map)

(where by a Triangular map we mean a map of the form T ( x , y ) = (x + h ( y ),y)).

Such a triangular map is invertible with inverse (x,y)»---(x-h(y),y)

f(x,y) = ((a.2 X + y) + a^ x + a^ y, k(a| x + — y) 2a|

+ b.x + buy)

So we may write it in the form :

2 2

f(x,y) = { (sx + ty) +a^ x + a^ y, k(sx + ty) + b^ x + b^ y)

s b. - ka 2a. b. - ka. provided that ^ - ka: (si"ce — = ^ )

5 5 2 5 5

and a^ b^ - a^ b^ ^ 0

This completes the proof of our theorem.

Remarks

1. We will see later how to allow for the case t=0 which strictly is not covered by the above.

2. The complicated calculation to verify the Jacobian Conjecture in this easy case gives an indication of why the Jacobian Conjecture is so hard.

The above procedure paves the way for describing the varieties of invertible polynomial maps. We will carry out similar calculation for the cases of degrees 3 and 4.

Theorem A polynomial map of degree - 3 whose Jacobian determinant is a non-zero constant is invertible and is of the form:

( x , y )

>— >

"here ty = = Rag '^ F ^ ' ^8 ^ 9 ^ ^9

The variety of all such maps is 7 dimensional.

Proof A typical origin-preserving polynomial map of degree - 3 is of the form:

3 3

f(x,y) = (a^x + ...+ agX + a^y, b^x + ... +bgX + b^y) and so the set of all such maps forms a space of dimension

18 and the condition det Jf = constant 0 defines 14 equations and one inequality in these coefficients.

As in the degree 2 case above, the leading terms are proportional and we compose on the left with the map:

g(x,y) = (x, kx - y) with k = — and &1 compose on the right with the map:

a?

h{x,y) = (x - g— y,y)

to get a map of the form:

3 2 3 2 2

gofoh = (m,j X + ir^ xy + m^y + m^x + m^xy + m^^y + mgX + m^y,

2 2

n^x + ngxy + n^y + UgX + n^y)

I

.

Note that the second component has no degree 3 terms

2

and the first component has no term in x y . For future reference we note that:

a^ = m,j ; = m^; a^ = m^ (coefficients of powers of x) and because the "linear part" of gofoh is g o (linear part of f)o h we must have a matrix equation:

k k

'■

mg mg

b g b g

3a

A

0

y

^2 ^8

leading to m^ = - +ag, ng = kag - bg and

"9 3a ( k a g - b g ) t ( k a g - b g ) .

Then the Jacobian determinant of gofoh is 3m^x + m^y + 2m^x + m^y +

2ngX + ngy + ng

m. 2mgxy +3m^y +m^x +2m^y tmg n^x + 2n_y t n,

Since the coefficient of x is zero we get 3m^ n^ = 0 and since (as in the last theorem) we may assume that m ^ ^ 0 we get Ug = 0. Taking this into account and equating the other coefficients in the determinant to zero we get;

2

X y : 2mg n^ = 3m.j n.^

2

xy : ng = 0

1

11 I

2

X : 3m,j rig = 2mg

xy: mg + 2m.^ = 2m^

2

y : mg rig + 2m^ = 3m^ rig

x: 2mg rig = m^ ng + 2mg n^

y : mg ng + 2mg n,^ = 2m,^ n^

We now show that n. = 0₅

Suppose by way of contradiction that n ^ ^ 0. Then m ^ = 0

2

(from the coefficient of xy ). 3

From y : Either mg = 0 or n^ = 0 2

From X y : If mg = 0, then n^ = 0 (since m^ ^ 0) If n^ = 0, then mg = 0 (since n^ 0) and so both mg = 0 and n^ = 0.

Hence from x y : m^ = 0

From y : mg ng = 0, then either mg = 0 or ng = 0. But mg = 0 iff Ug = 0 (from x^)

and so we have ng = 0 and mg = 0 anyway.

But then mg= 0 (from x). But this is a contradiction, since | 1 mg H g - Mg Ug 0 and so we must have n^ = 0 as claimed.

2 2

Since n^ = 0, it follows (from the coefficients of x y and x ) that n^ = ng = 0.

and then = m ^ = mg = m.^ = 0. Thus the map gofoh is of the form:

3 2

{x,y )i— > (m,|X t m ^ x + m g X + m g y , r^x)

and so (as in the degree 2 case) f can be written as a composite :

(Linear map) o (Triangular map) o (Linear map) which is clearly invertible.

Since n g = 0, it follows from the formula for n g that 3a. (kag - bg) = a^ (ka^ - bg )

We must have ka g - 7^ 0 , otherwise ka g - b g = 0 (since a^ ^ 0) and this would give |jfj= 0.

3a (kag - b g )

Therefore a. = — — ,---r-— — _{2 kag - bg} and a. is determined by:₂ (a,|, b.j, a g , b g , a g , bg).

Using the formulae for m^ and n ^ in terms of a^ and b^^ we get

3 2 ^9 ^2 ^8

f = (x, — X - y ) o (a. X + a^ x + a. x + --- y,

^1 1 5 8

b, 3g ^ 2

- bg)*) ° <* + 357 Y' Y).

Note that f is determined by:

h,' S' S' S' S' S'

From above:

^ a a a^

f(x,y) = ( ( ai^ X + ~ ~ y ' 7 + ( a= x + y)^+ a^x + a^y,

k

1 . 1 = = 2

T 7 3 2 2 5 2

k ( a ^ X + Y — y) + k{a^ X + y) + bgX + b^y) 3 a ?

So we may write it in the form:

3 2

f(x,y) = ((SgX + tg y) t (s^x + t^y) + ag x + y

k(.Sg X + tgy)^ + k (s^x + t^y)^ + bgX + b ^y for some s,j, t,^, S g , tg e (E provided that

a 1 S. ka g b g

— = — = (since from the formula for n^,

t l ^ 2 k a g - b g

3a. ka. - b

- ) and a. b. - a. b. ^ 0

S - b g

This completes the proof of our theorem.

Remark We will see later how to allow for the case t^ = 0 which strictly is not covered by the above.

We now look at the next hardest case: degree ^ 4. Here there are two possible kinds of invertible polynomial maps and we get the following result:

Jacobian determinant is a non-zero constant is invertible. Any such map is of one of the following types:

1) (x,y)»— > ((SgX + tgy (SgX + tgy)^ + (s^x + t^y)^

+ a^gX + a^^y, k(SgX + tgy)^ + kfSgX + t.yl^f

k(s^x + t^y)2+ b^gX + b^^y)

where s. s. s. ka.^ - b.^

t; = Ü = = to,7 - "b,; and ^13 b,4 ^ b,3

2) The composite of a pair of invertible polynomial maps of degree 2.

The maps of type 1) form a variety of dimension 8. Maps of type 2) also form a variety of dimension 8.

Proof A typical origin-preserving polynomial map of degree - 4 is:

f(x,y) = { a ^ + .... + a^gX t a^^y, b^x^t ... + b^gX + b^^y) and so the set of all such maps forms a space (over C ) of dimension 28 and the condition det Jf = constant ^ 0 defines 27 equations and one inequality in these coefficients.

As in the easier two cases, the highest order terms in the Jacobian determinant lead to the conclusion that the highest order terms in the two components are proportional and we may compose on the left with the map

We compose on the right with the map

h(x,y) = (x - Y , y) to get gofoh (x,y) =

4 2 2

(m^x + m_x y + + m^^x + m^^y, n^x^t n y+ ....fn^^x + ,y4

As in the last case, m^ = 0 and there are no terms of highest degree in the second component . For future reference we note that:

^10 ^10

a and because the "linear part" of gofoh is

g o (linear part of f ) o h we have a matrix equation

leading to: _{”^13 ^13}

m 14 '^2 ^13

4a-^13 ^1

+ a 14

13 13

and _{n..■| 4} - -.--.yr --- ^2 ^13 ^1 , ^2 ^13 . ^14 ^14- +

---4a 4a ₁ 14

The Jacobian determinant of gofoh is

3 2

4m- X + 2m^ xy +

3n^ X + 2n_- xy +

2m^ X y +

n^ X +

ï

5 ÿ

and from the coefficient of x we deduce n^ = 0. Using this -'g t fact and calculating all the other coefficients we deduce the .*

following 19 equations: |

X y; 4m^ n^ = 3m^ n^

xy^: m^ n^^ + 4m^ ^io “ ^”^3 ^12 3m^ n^

y'^: 2m^ ^12 " ^"^5 ^11 ^8

x^: 4m, "i4+ 3mg n,, = 2m ^ n,^ + 3m,,

t

x^y^: 4m, rig = 3ra^ rig f

2 3

X y : m^ n^ = 6m^ n^

xy : m^ ng = Gm^ n^

y^: 3m^ n^ = 4m^ Ug

4

X : 4m^ n^^ = 3m^ n^

X y : 4m^ n^ ^ + Sm^ ng = 2m^ n^g + 3mg n^ 4

2 2

X y : 3m^ n^ + m^ Ug = 2m^ n^g + n^ |

M

X y: 6mg ^ + 4m^g = 2m^ n ^ ^ + 4mg n^g + ^^12 ^6

xy : 2mg 4m^ ^ + Gm^gOg + = 3m^ _{4 13} + mQn^^+_{8 1 1} 6iHQn_{9 10 ':}

y : ^ 4^ i 4 + 2nig ^ + 3m^ ^ n^ = 4m^ n^g + 3mg + 2m ^ 2^j

X : 3mg + 2m^g = m^ n^g + 2m^^ n^g + 3m^^ n^

xy: m^ ^14 + ^m^g ^ 3 "g "^g ^13 ^^12 ^10

y : nig + 2m^ ^ n^ ^ + 3m^g = 3mg n^g + 2m^2 ^ g

x: 2m^g n ^ ^ g ^11 ^13 ^ ^^14 ^10

y: + 2m^g = Zm^g n^g +

We shall show that all the degree 3 terms in the second component vanish. We start with:

Lemma The Coefficient n^ = 0

--- D

Proof of Lemma: Assume by way of contradiction that 7^ 0 Since = a^ 7^ 0, then from

3 2 X y :

4 xy :

ng =

m

^■"4"6 4m-13mg \

= 6mg ^ 3*4 "e\ (4mi

Hence = 6mg

Thus nig = 0 and either nig = 0 or = 0 We look at these two cases separately.

Case a) If nig = 0, then from x^y: n^ = 0, 2 3

from X y : mg n^ = Gm^ n^ , then m^ = 0

5

from y : either m^ = 0 or n^ = 0

3 2

from X y : m^ = 0 iff n^ = 0 and so both these vanish and we have

m 3= m4 = mg = ng = ng = 0

3 ?

Case b) If m^ = 0, then from x y : Ug = 0, from y ^ ; either mu = ₅ 0 or n. = ₈ 0

2 3

If mg = 0, then (from x y ) either mg = 0 (and we are back to case a)) or ng - 0. But if Ug = 0 then (from x^ y) mg = 0 and we are back to case a) anyway.

So in either case we have mg , m^ , mg , ng and n^ = 0

and we have exhausted the information of the degree 5 terms. So we proceed with the lower degree terms.

3mv ng

" 8 6

and we may cross-multiply and cancel to get m? n^2= nig n^ ^

Now from _{x V =} mg == 0

À

2

But from xy : 4m^y 2 == nig and so m^ ^12“ ^

Hence either m^ = 0 or n^2 = 0 1' We treat these two cases separately (Cases c) and d ) )

Case c) m ^ = 0, then we have n ^ ^ = 0 (from x^ ).

So Case C^)leads to a contradiction.

Case C g ): n ^ ^ = 0

Note that we also have m^, m ^ , m^, m ^ , m^, ng, n^ and n^^= 0

3 2

From X : m^^ = 0 and from x : m ^ ^ = 0, but this too leads to a zero determinant and so Case C2) is also impossible. Now we return to case d).

-a

4 Then m_ = m. = mu = m_ = m^ = n ^ = n -, = n . - = 0 _{3 4 5 7 9 8 9 11} -From x^y; 4m^ ^12 ~ 3mg n^

X^: 4m, n,^ = 3m,,

and we may cross-multiply and cancel to get m^ 1 n^2 = nig n^^

2

From y : mg n^^ + 2m^^ n^2 = 0 f then 3mg n^^= 0 and either mg = 0 or n^^= 0 (Call these cases C ^ ) and C2))

Case C ^) mg = 0, then n^ 2 “ 0 (from x^y) 2

and m^2 ” 0 (from x y). Then either m^^= 0 or n ^ ^ = 0 3

(from y) and from x : m^^ = 0 iff n^^ = 0 and so both are zero 2

Then from x : m^^ n^ = 0 and then m^^ = 0 and so n^^ ~*^14 “ ^ ' which is impossible, since det Jf = m^g n ^ ^ - m^^ n^g ^ 0 •

C a s e d) n ^2 = 0

N o t e t ha t we a l s o h a v e iDg, , nig, nig, rig, a nd n g = 0

3 2

F r o m X y; m^ = 0 a n d f r o m y : m^2 " 11= 0

an d so e ither n ^ ^ = 0 or m ^ 2= 0

If n ^ ^ = 0, from x^: = 0 a n d we are r e d u c e d to c a s e c). On 2

the o t h e r hand, if m ^ 2 - 0 f t h e n from x y: n^^ = 0 a n d

e i t h e r m^ = 0 or n ^ ^ = 0.

B ut f r o m x^: m ^ = 0 iff n ^ ^ = 0 a n d so w e a r e r e d u c e d to

c a s e c) anyway. Thus n g = 0. T his c o m p l e t e s the p r o o f of

t he lemma.

C o r o l l a r y of l e m m a W e a l s o have: n g = n g = 0 a n d

4 2 2

g o f o h = (m^ x + m^ x y + ... + m^g x + m ^^y,

"10 "11 + "12 + "i3 X + "i4 y)

3

P r o o f of c o r o l l a r y : F r o m x y : ng = 0 and 4

f r o m X y: ng = 0

a n d so as c l a i m e d all t h e d e g r e e 3 ter m s in t h e s e c o n d

c o m p o n e n t vanish.

T a k i n g a c c o u n t of t h i s r e s u l t w e m a y r e w r i t e the

e q u a t i o n s above:

x"^: 4m^ n^^ = 0 a n d so n^^ = 0

3

X y : 2m^ n^ 2 = ^g n^g

3

xy : 5 ^10 ^3 ^12

3

X : 2m^ n^g

2

X y: SiHg n^2 = n^g + 2mg g

2

xy : 2mg +

4

3

y^: m^ + 2mg n^2 = ^m^ n^g

x^:

3

xy: n,^ + 2m,^ n,^ fflg n,3 + 2m,g n,g

y : mg + 2m^^ 2 =

3

x: Zrn^g ^1 1 "13 Zm^^ "10

y: ^11 "14 + Zrn^g Zm^2 ^13

2 2

From the coefficient of x y we get two cases

" 10 = 0 andn^g/^ 0

These correspond to the two types of map in our theorem.

1

Case 1

•3

If n^g = 0 , then from x : n^ ^ == 0 and (since the determinant is a non-zero constant) we must have n^^^ 0.

From X y: n^ ^ “ 0 (since m^ = a^ ^ 0)

From X y: m^ n^g = 0 — .m^ = 0

2

xy : m^ n^g = 0 '.... — ^ m^ = 0

y^: mg n^g = 0 ^ = 0

y: 2

X : m^ n^g = 0 -A m^ = 0

xy; mg n^g = 0 mg = 0

y : iHg n,3 = 0 ^ ">9 = 0

X: m ^^ n^g = 0 - \ m^^ = 0

12 ^13 ° = = ^ ^12

and so in this case

4 3 2

gofoh (x,y) = (m^ x + m^ x + rn^gX + rn^gX + m^^y, n^gX)

Hence f can be written as a composite:

(linear map) o (triangular map) o (linear map) which is clearly invertible.

since " 14 = 0, it follov ^1

1 ®14 &2 "

4a _{“ ^14}

pL ,

a, ®13 ^13

is determined by a ^ , b^ , a Using the formulae for m^ ,

gofoh = (a, 4 X + a_x_o 3 + a^g

a,3 - a.

^13 X)

*=1

— x-y)

f = (x, O (a^

4a, ^14 " ^2 A r.

a , 3 ^

and a2

13 4a^

(X + ^ y, y)

4 3 2

: + agX + a^gX + a^gX +

bi a^g a^ b^g

--- X ) o

^1' ^1' ^6' ^10' ^13' ^13' ^14' ^14*

This result indicates that "locally" the variety of degree ^ 4 invertible polynomial maps of type 1 ) is 8 dimensional and we will see later that this, is true globally.

^^1^14 *2^13

..H

Note that f is determined by:

A: '-j_'S;

From above:

1

1 1 3 a

14 „ , ..,4. ,_I_. , 3^

4a-4a" 3 4 1 1 2 ^2 *^10 51 1

2 *in 2 *'•

(&10 X + — ---- y) + a^gX + a^^y, X

1

1 1 3

% a^ 4 g a^ ag g

k(a^ X + — g y) + k (ag X + — ^ — y) +

4ai

I 1 ^ ^2

9 a^ a - 9

k(aio X + 4ay" Y ) + b ^ g X + b.j^y)

and we may rewrite this as

f{x,y) = {(Sg X + tgy)^+ (s^x + tgy^ + (s^x + t^y)

+ a^gX + a^^y, k(SgX + tgy)^ + kfSgX + tgy)^

+ k (s^x + t^y)^ + b^g X + b^^ y)

Provided that

s - s„ s^ ka-^ - b-q 4a kà-_ - b-^ _ L = _2 = _3 ^ _ _ y --- U i (since ^ ^ )

t , t 2 t 3 k a , 4 - b , 4 A k * 1 4 A 4

%

4

I

and a,3 b,^ - a,^ b,g ^ 0

This is the form given in the theorem. We will see later how to allow for the case t^ = 0 which strictly is not covered by the above.

Case 2 This is the case n^g^ 0, m^ = 0. Recall that we also have: n g , n^ , n^ and n g = 0 . We first show:

Lemma The coefficient m^ = 0 Proof : Suppose that m^ ^ 0

3

From the coefficient of x y we get "12 ^ Then using the

3 3

coefficients of x y and xy and multiplying we get

2 2

2m^ mg n^2 = Zmg m^ n^g |

0 9 ^5 ^12 2 2

and hence m, n,^ = n,^ or — = ( — )

1 “ 10

"12

(say) where s = 7^ 0.

"10

From the same two coefficients we also get: ^ 3 "10 ^12 ^ ^5 "10 "12

2 2 2

and so m^ = 4m. m^ = 4m. s_{3 1 5 1}

So our map gofoh is of the form:

(x,y)H-f(m^(x^ + sy^)Z + lower terms, n^g(x^Fsy^) + n^gX + n^^y) But composing this map on the left with the invertible map:

m- 2 2 (x,y) »--- > (x - — ^ y , y) gives a map of the form:

Ao

I

2 2 r

{ x , y ) i— ^polynomial of degree - 3, n^g(x +sy ) + n^gX + n^^y) '4

From the earlier theorem about invertible maps of degree - 3, the first component can have no terms of degree 3 But from the earlier theorem of degree - 2 , this map could

only be invertible if s = 0 which gives us our contradiction . | A This completes the proof of the lemma.

Since mg = 0, we deduce from:

3

X y : n^2 ~ 0

3

xy : mg = 0

2

X y: mu = 0

2

xy : m^ = 0

4 3 2 2 2

Hence gofoh = (mpx + m^x + m^ x y + m^^x + m^^xy +

+ n,gX + n^^x + n,^y)

The equations deduced from the Jacobian become

x^ : 2m^ n^^ = m^ n^g 1

2

X ; 3iïïg n^g + 2m^ ^ n^g 3

Zrn^g n ^ ^ ^1 1 "i3 ■*■ Zm^^ "i o ^

y: ”^11 ^14 " Zm^2 >^13 *... %

From these we deduce: Zm- n - . From 1 : m-- = —

---"io

” 7 "14 ” 1 "14 2= ” 12 = - 2^7^ = - 7

-^10

Now ^ 0 otherwise we have m^ ” ^^12 ” ^

and from 3: m.^ ^ = 0 and then from 4: m^^ = 0 which would give a zero determinant.

From 5 : m_{11 “} 2m^2 _" "13 _" Zm^ _^2^13 ^14

"14 10

m n "^14 ^10 "^1 "d3 , "^14 ^10 4: m^g = + = ---2— + -

---Z"i4 "14 "10 ^14

^7 "13 2m^^ n^g 2m^ n^g 4m^ n^g 3 . m ^ = - .... + --- = — — --- +

Z"i4 3n^^ 3n^g 3n^g ^^1 ^13

"10

28

I

(x,y) , > ( m %2 + :

' "10 "lO s

mi n^g "1 0 , 2 , "l3 ^14 , "

2— ■*■ --- ) X + ---g--- xy +

"10 ^14 "10

m.j n 14 2 2

2— y + m^g X + m^^y, n^g x + n^gX + n^^y)

"10 *

2 1

= (m (x^ + ^ y)2 + .^1 ^1.3 ^ !2ll2lÊ)xZ I

A o A o n,, I

j % 2m,| n^g

+ xy + m^gX + m^^y, "fo

2 ^14

"10 + 7 7 - y) + "13%)'

and this is the composition:

^1 2 ^ 1 4 ^ 1 0 2 2

(x + -2 y ' y) o (— n ^ + m^gX + m^^y, n^gX + n^gX+ n^^y)

"10 14

From the result on invertible maps of degree - 2 the "right

hand" map is invertible. Hence, as the theorem stated: in s case 2) the map is the composite of a pair of invertible

This result indicates that the variety of degree 4 invertible polynomial maps in this second case is 8 dimensional "locally". We shall see later that this is true globally.

This completes the proof of the theorem.

Remark

Proof

Let f , P^ and be as in the statement of the theorem. If we look at the terms of highest degree in the determinant of Jf we get: D» P^ . D q - D q^ . ID, P^ = 0 . _{^ x r y ^ s x ^ s y r} In other words the Jacobian determinant:

^ ( x , y )

From the general solution of this partial differential equation we deduce that there is a functional dependence between P^ and cy . That is, for some (differentiable) function F we have F( P^ , q^ ) = 0

Since P ^ , q^ are polynomials, F is also a polynomial. General Results on Maps with Constant Jacobians

In the calculations in the last chapter it was clear that the highest order terms of a polynomial map with constant Jacobian can be easily dealt with by looking as the highest order terms in the Jacobian determinant.

Theorem (Cf [Vi] pg 415)

I Let f(x,y) = (P^ + P^_^+ q^ + q^_^ + ) be | a non-linear polynomial map with constant Jacobian. If P_{r , 7} Qg ^re the leading terms and are homogeneous polynomials

We now look at those terms of F which give us terms in X and y of degree k. These must vanish as polynomials in x and y and so we have

ir + js k

Let be the term with non-zero coefficient with highest power of P^. Divide throughout by this and we find; Lemma : We get a polynomial in t where

Hence after division a typical term is

%

a . . — where a = j - j ,b = i - i

h o o

L

t and the lemma follows.

qr/d J

t = ----:— and d = hcf (r,s) pS/d

r

i 1 ^

Proof of Lemma: Since P^o q^o contains the highest power % of P^ it also contains the lowest power of .

1

Now since ri^ + sj^ = ri + sj = k we have

j - j r(i - i^) + s {j - j^) = 0 and so | =

o

In its lowest form the left-hand fraction is --44_s/d

The polynomial in the lemma factors into linear factors over (H say;

e{t-a ^ ) (t-^ ) .... (t-a^ )

and so the polynomial factors into a product of factors of the form

-

-i - r *

As polynomials in x and y one of these must be zero and so we have;

as in the statement of the theorem. Remark

In the last chapter when we investigated maps of degree 2, 3 , 4 we showed that if the two components had the same degree then the leading terms were proportional. This .is just the case r = s of the above theorem.

This last result allows one to attack the Jacobian Conjecture from a different direction, as the following corollary indicates.

Corollary

To prove the Jacobian Conjecture it is enough to prove that for any polynomial map with constant non-zero Jacobian

f(x,y) = (p(x,y), q(x,y))

Proof of Corollary

We have (say) the degree of q dividing the degree of P Then in the above theorem r/s = n is an integer and so the leading term of P is a multiple of a power of the leading term of q. So p-cq^ has degree ^ degree P and we may write

f = (x + cy" , y ) o f^ with

f^ = (p - c f , q)

Then degree f^^ldegree f (If degree p = degree q we need to apply this method again to get strict inequality) and f ^ still has a constant non-zero Jacobian. Thus the Jacobian Conjecture will follow by induction on the degree.

This completes the proof of the corollary.

Remark 1)

The proof of the corollary suggests that any invertible polynomial map can be written as a composite of maps of the form (x,y)#----^ (x +Cy^, y ) and linear maps or equivalently as the composite of linear maps with what, in the last chapter, we called "triangular maps"

(x,y)i ---- ^ ( X + h(y), y ) for a polynomial h.

This result is in fact true and is one of the oldest results in this part of the subject. It was proved by Jung in 1 942 [J ] and has been reproved several times since (eg. Van der Kulk [V] 1953).

Remark 2 )

The divisibility condition in the corollary follows in some special cases from a theorem of Magnus [M] 1 955; "If the degree of two polynomials are coprime and their Jacobian is constant then this constant is zero"

This proves the Jacobian Conjecture for the case of prime degree. A generalisation of Magnus's theorem by Nakai and Baba [NB] 1977 proves it for degrees which are twice a prime.

Remark 3 )

It is not obvious that the divisibility in this corollary holds for any invertible map. In fact Segre [S] (1956) stated the following lemma as part of an attempt to prove the Jacobian Conjecture:

If f, g e C [t] are non-constant polynomials and generate the polynomial algebra then degree f divides degree g or vice-versa. This result implies that the divisibility condition holds, since if (x,y)»— >(p(x,y), q(x,y)) is inver­ tible with inverse (x,y) »---- > (r(x,y), s(x,y)) then X = r (p (x,y), q (x,y)) and we may replace y by a suitable ux in the identity and use Segre's lemma.

The lemma was proved in 1975 by Abhyankar and Moh [AM] See [BCW] pg 299 for a history of its attempted proof. The paper [AM] also shows that the lemma implies Jung's theorem

We now show how use of these results can save some of the calculation of the last chapter.

A) The degree is prime (in particular degree = 2 or 3) or the map cannot be written as a composite of lower degree m a p s .

From the theorem above and Magnus ' s theorem, either the two components have the same degree - in which case we can compose with a linear map to reduce the degree of one of them, or one of the components (say the second) is linear. In the case r = 3 considered in the last chapter this avoids the rather tricky proof that n^ = 0 which enabled us to eliminate the second order terms from the second component.

In the case of a general prime our map is now of the form;

(x,y)i ^(c^ (sx + ty)^ + ...., sx + ty) and this is the composition:

(x + c^ y^ , y ) o (P^_^+ ..., sx + ty) and by the same argument:

Cg (sx + ty)^“ ^

and so on.

f

Thus our map is:

One may then deduce the standard form for invertible polynomial maps of prime degree or for irreducible maps i

{x,y)i ^ (( s^_^x + + ... + (s.j x + t^ y ) ^ + ax + by.

k(s^_^ X + t^_^y) + ... +k(s^ X + t^ y) + cx + dy)

where Sr_i _{t;::; - - t; - kb - a}ka - c

B) The degree 4 case.

In this case we may compose with a linear map to get the second component with d e g r e e ^ 4. If this second compon­ ent is linear then we are reduced to last case.

From Magnus ' s theorem we deduce that the degree of the second component cannot be three and this is what we proved "the hard way" in the lemma in the last chapter in which we showed that n^ = 0. The only other possibility is that this second component has degree 2. So we have a map;

(x,y)*— + P3 + + P, , q% + q, )

2

and from the theorem above, = cq^ for some ceC and so our map is a composite

(x,y)<--->(x + cy^ , y) o ( P^ + P^ + P.J , q^ + ) and (as before) since the "right-hand" map is invertible we must have P^ = 0 and then our map is the composite of two degree 2 maps as we showed earlier.

To give an indication of how more complicated cases work we briefly consider:

C) The degree 6 case*

Assume that the first component (say) of our map has degree 6. Then the possibilities are:

1) Our map is of the form

(x,y) »— ->(degree 6 polynomial, degree 6 polynomial) As before, we may compose with a linear map to reduce the degree of the second component and get one of the following c a s e s .

2) We have a map:

(x,y)i— ^(degree 6 polynomial, degree 1 polynomial) This case has been handled in A) above.

3) The map is :

(x,y)«— » (degree 6 polynomial, degree 3 polynomial) = (Pg + ... .. qg + ... )

2

Then Pg = cq^ from our theorem and our map is a composite: 2

(x + cy , y ) o (Pg+ ...., qg + ... )•

From the results about invertible maps of degrees 5 and 4 the right-hand map must in fact be of degree 3 and so our map is a composite of a degree 2 with a degree 3 map.

4) The map is

(x,y) »---- > (degree 6 polynomial, degree 2 polynomial) = (Pg + ... qg + ...)

(x,y) I-y (X + cy3, y) o (Pg + + ... , 9% + ) ;

Since the right-hand map is invertible we have = 0 and w

2 "

then P 4 = C.J q 2 for some € (Î. and so our map is: ï

3 2

(x,y) ^ (x + cy- , y) o (x + c^y , y) o (P^ + q2 + q.j ) a; ri

3 2 '

(x + cy + c.jy , y) o ( P^ + ... q^ + q^)

and since the right hand map is invertible it follows that = 0 and our map is a composite of a map of degree 3 with a map of degree 2.

Remark : From Magnus's theorem the degree of the second component cannot be 5. From the generalization of Magnus's theorem [NB] it cannot be 4.

We may use the method outlined in the last case to give a new proof of Jung's theorem from Segre's lemma

(different from that in [AM]).

Theorem: Any invertible polynomial map is a composite of linear maps and triangular maps. (Jung [J]).

Proof : We call an "upper triangular map" one of the form

(x,y) I > (x + h(y), y )

(The Jacobian matrix of such a transformation is an "upper ’ triangular matrix"). A "lower triangular map" is one of the

Given an invertible map f:

(x,y) I > (P^ + ... qg + )

then by Segre ' s lemma (say) s divides r (if not then r divides s and we may compose with the linear map:

(x,y)i— >-(y,x)). Then from the first theorem in this chapter Pj. = cq^/^ and f is the composition:

(x + cyf/s ^ y ) o (P + ... q + ) fl/s

Either r-<2^s or s divides r^ and P _{1 1 r^} = c^ q _{I ^s} and f is the composite;

r/s

(x + cy + c^y , y) o (P^^ + q^ + ...

Repeating this process gives us a composite of the form; (x + h(y),y) o (P^ + , q^ +

with f ^ s and hence f dividing s. Use the same method to write f as;

(x + h(y),y) o (x, y + k(x)) o (P^ + ... q^ + ... ) with s f .

We continue this process until we are left.with a linear map and our result is complete.

The above proof shows that we have a slightly stronger

version of Jung's theorem: Any invertible polynomial map can y be written as a product of the form: | Vd ° o Ug o ... ° " k ° o M I

'1

with U . an "upper triangular map", L. "lower triangular map" | I and M a linear map. Moreover we may arrange that degree

If we start with a product with a linear map "in the | middle" our algorithm produces a new product with the linear

map at the end. .

2 2

eg (x + y , y) o (x, y + x) o (x + 2y , y )

(x + y^ , y) o (x + 2y^, y + x + 2 y ^ )

2 2

(x + y ,y) o (x + y,y) o (-y, x + y + 2y )

2 2

(x + y + y,y) o (x,y + 2x - x) o (-y,x)

We can then show:

Theorem: The degree of our map will be the product of the ^ degrees of the triangular maps.

Proof : Clearly we need only calculate the degree of o L ^ o ... o U^oL^

We will assume that the "final map" L»^ is not the

identity (ie the product does not finish with an upper f triangular) - otherwise conjugate with the map: (x,y)»—>(y,x)

which leaves all the degrees unchanged but reverses upper and lower triangular maps.

Lemma : The degree of a product of alternate upper and lower

triangular maps starting with a lower triangular map is the fi degree of a power of x in the second component and the

case the degree of the product is the product of the degrees of the components.

Proof of the lemma: This is by induction on the number of 1 maps in the product. So we consider the product U o (p,q)

with U the upper triangular map: (x,y)»-- > (x + u(y),y) and (p,q) a map whose degree is the degree of a power of x in the second component.

Then this map is (x,y)i— »(p + u(q),q) and the result follows provided that either degree u or that p,q have different degrees (otherwise the highest order terms of p and u(q) might cancel one another).

This latter condition will be satisfied provided that no two adjacent components are linear. The other part of the inductive step is to consider products of the form L o (p,q) and the argument is similar.

This completes the proof of the lemma.

The theorem follows from the lemma since the form of our decomposition only allows successive linear maps if the whole product is linear.

Now we look at the composition f o g :

og = ^3^+-] o ^k+J ° ... o o (L^ ) o where is linear map.

So f o g = o o ... o U % o ( ) o ^-j p ... o (L^ ) o

1* If is non-linear then no collapse can occur and degree f o g = degree f . degree g (Because the lemma in the above theorem still holds if there is a single linear map in the product)

2. If I^ is not present (if is linear include it in ) then we have a pair of upper triangular maps coming together and some "collapse" can occur.

a) If degree ^ degree then the degree of o is the higher of the two degrees and the degree of fog is lower than degree f . degree g by a factor of the lower of these degrees.

Note that degree fog divides degree f . degree g .

b ) If degree = degree U^^^the situation is harder.

- 1

Firstly and U^^^might cancel completely ie = ^k+1 and further collapse of and might occur.

Secondly, degree U ^ o can be anything from 1 to degree and in this case degree fog will not necessarily divide degree f . degree g

Chapter 4 4

i

Standard forms for invertible polynomial maps from C to itself

.1

1

In this chapter, we develop some standard forms for

the invertible polynomial maps of different degrees ' considered in chapters 2 and 3.

For any polynomial maps f, g we have:

(linear part of f ) o (linear part of g) = linear part of fog and hence any invertible polynomial map f with linear part M can be written as M o (M f ) I the composite of a linear map with a map whose linear part is the identity. Thus it is enough to write down standard forms for maps with linear part

Degree

Taking the standard form from chapter 2 and restrict­ ing it to the case when the linear part is the identity we get the following standard form for such maps:

i x , y ) i y ((sx + ty)^ + X, " (s/t)(sx + ty)^+ y)

However, this does not allow for the case t = 0 for which we may have a map of the form:

2

(x,y)i ^ (x, ex + y) for some e e C

I

Again, we consider maps which preserve the origin and whose linear part is the identity. Then from chapter 2, we can determine that such a map is of the form:

3 2

(x,y) j ^ ((Sg X + tg y) + (s^ x + t^ y) + x,

k f S g X + t^ y)^ + k(s^ X + t^ y )^+ y) (x,y) I ^( t( s x + ty)^ + X, - s (sx + ty)^+ y)

for any s, t e CD (of course the s, t will in general be different from before). Note that any choice of s, t will give such an invertible polynomial map, but that several different choices will give the same map.

In fact, if our map is of the form:

2 9 2 2

(X ,y )I-— -y (a^ X + ag xy + y + x, b^ x + b^xy + b^ y + y) 1

3 2

then t = a^ and - st = b^

and so if a^ ^ 0, the number t can be any of the three different cube roots of a ^ and then the number s is determined.

If a^ = 0, then t = 0 and from - s^ = b^ 0, one can take s to be any of the three different cube roots of -b^.

Thus each invertible polynomial map of degree - 2 which preserves the origin and whose linear part is the

2

identity corresponds to exactly three pairs (s,t) e CD .

45 1

^2 ^ 1

with - 3T- = - IT- = k

^2

are determined.

If a^ = 0, then a^ = 0 also and = 0

i

'i However this does not allow for the case where both of the

t's are zero for which we may have a map of the form: (x,y)»-- > (x, e^ + e^ x^ + y)

for some e. , e ^ e (E. We can combine these into a single f formula in a way similar to the case ( degree - 2 ) by: % (x,y)i— ^(tg (Sg X + tg y)^ + t^(s^ x + t^ y)^ + x, J

- S2 (S2 X + tg y)^ - s^(s^ X + t^ y)2 + y)

where now the ratios S2:t-2 and s^:t^ are the same or undefined. Equivalently S2 t = s^ t2 .

As in the case ( degree - 2 ) , any suitable values of the numbers s^ , t.^ , S2 r 12 r will give such an invertible polynomial map. Again several different choices lead to the same map.

In fact if our map is of the form:

(x,y)i— ^ (a^ x^ + ... + a^ y^ + a^ x^ + + a^ y^ + x, b^ x^ + ... + b^ y^ + b^ x^ + .... + b^ y^ + y)

4 3

Then t2 = and -S2 t2 = b^

3 2

t^ = a^ and -s ^ t^ = b^

■Â and so if a^ /: 0 and a^ ^ 0, the number t2 can be any of the

Then from -s^ = b^, - = b ^ , we can take to be any fourth root of - b^ and any cube root of - b ^ . Thus each invertible polynomial map of degree 3 which preserves the origin and whose linear part is the identity corresponds to exactly twelve 4 - tuples:

(s^, t ^ , Sg, t p ) B d ^

Degree - 4

From chapter 2 , we must consider the following two cases :

Case 1 : An irreducible invertible polynomial map which preserves the origin and whose linear part is the identity is of the form:

(x,y)i— > ((Sg X + tg y )^ + (Sg x + tg y )^ + (s^ x + t. y )^ + x,

k(s^ X + t^ y)"^ +k(s2X + t2 y ) ^ + k(s^ x + t^ y ) ^ f y)

with - Sg/tg = ~ ^ k

However this does not allow for the case where the t's are zero for which we may have a map of the form:

4 3 2

(x,y)-- ^ (x, eg X + e2 X + e^ x + y)

for some e ^ , 62 ^ e^ e C. We can combine these into a single formula in a way similar to the other cases by:

( X , y ) •-— ^ ( tg (SgX + t g y ) ^ + t 2 (s2X + 1 3 (s^x + t ^ y ) ^ + x.

'I

- S g ( S g X + tg y ) 4 - (S2 X + t2 y)^ - S^(s^ x + t^ y)^ + y) |

where now the ratios Sg:tg , ®2*^2 ^3

Si t] = Sg

and tg = Sg

As in the earlier cases, any suitable values of the numbers , t^ , s 2^ f Sg, tg will give such an invertible polynomial map. Again several different choices lead to the same map.

In fact arguing as above, the pair (s g, t g ) is determined by the map up to multiplication by a fifth root of unity, the pair (s 2 , ^ ) up to multiplication by a fourth root and (s^ , t^ ) up to multiplication by a cube root.

So each irreducible invertible polynomial map of degree - 4 which preserves the origin and whose linear part is the identity corresponds to exactly sixty 6-tuples:

s^, t^, S 2 , t 2 , Sg, tg) S

Case 2

We shall deal with these as a special case of the following important uniqueness theorem.

Theorem; Any reducible invertible polynomial map f whose linear part is the identity can be written as a product of irreducible maps whose linear parts are all the identity.

If the product of the degrees of the maps in the decomposition is the degree of f then this decomposition is unique.

Proof : We start by writing f as a decomposition of alter­ nate upper and lower triangular maps as in Chapter 3.

If f = (p,q) then there are several cases.

1 ) degree p ^ degree q. Then the decomposition starts with a non-linear upper triangular map and we may write f = U o g^ where g = (p^, q) with degree

p^ degree q.

Then by its construction there is a unique map U with this property. Note, however, that U will not in general have linear part the identity.

2 ) degree p ^ degree q. The decomposition starts with a non-linear lower triangular map and we have f = L o g^ where g^ = (p, q ^) with degree p ^ degree q ^ . As before L is uniquely determined by this property.

I